# Tag Info

18

The identity $$V = K \frac{dV}{dt}$$ is only guaranteed with a constant $K$ if your assumptions actually hold. The first identity $V=RI$ only holds for a resistor, while the other holds for a capacitor. So in this sense, the letters $V,I$ in these equations mean something else. In one of them, it's the current through (or voltage on) a particular resistor, ...

6

This is true but strictly limited to RC circuits without external sources: that is, a resistor hooked up to a capacitor with nothing else in between. In that case, $V$ is indeed proportional to $\dot V$, with a crucial minus sign in between: $$\dot V=-\frac1\tau V,$$ where $\tau>0$ is some constant. This equation implies that $V(t)=V(0)e^{-t/\tau}$, ...

5

Generally, no. You've written two equations. The first relates voltage and current for an isolated resistor. The second relates voltage and current for an isolated capacitor. Given only those two expressions, there's no reason at all that they can or should be combined. That is, you have no circuit, only isolated components. There are principles for ...

5

Does that mean when I apply a voltage, the current will be infinite large? No, not even in the context of ideal circuit theory. It's a bit subtle since we're using phasor voltages and currents and that requires a couple of assumptions to hold in order to be valid. When those assumptions don't hold, we have to see what the 'infinity' (division by ...

3

Typically, with inductors, we use the complex impedance, $Z=i\omega L$ with current frequency $\omega$ and inductance $L$, for the voltage: $$V_{ind}=IZ,\quad V_{rms}=I_{rms}|Z|$$ where the left equation is the voltage of the inductor and the right equation the root-mean-square. Surely, however, what goes on inside the inductor doesn't matter, it only is ...

3

It is quite easy to use a step-up converter to generate almost any voltage. The question is "voltage at what current". The power a battery can deliver is finite (power = voltage times current), but you can convert voltages in many different ways. The most obvious is an oscillator (inverter) followed by a transformer and a rectifier, but there are more ...

3

The $KV$ in $30KV$ refers to the motor velocity constant $K_v$. It does not refer to kilovolt $kV$, cf. comments by John Rennie.

3

The simple reason is that you have only passive, linear components in your circuit; when you push with a certain voltage, you get a certain current response. If the voltage cycles a certain number of times per second, that's how often the current must respond. Think about it the other way - if you _didn't_have the same frequency, what response would make ...

3

This is solved using start-delta transformation and Delta-Star transformation like shown in the below picture when you rewrite the circuit diagram it looks like this and you can notice delta network and star network in it. Applying proper transformation reduces the complex network to simple network, then you can solve simple series/parallel combination ...

3

It's not a dumb question. The electrons at the negative side of the battery have energy, and will look for "any path" to get rid of that energy - move in the direction of an electric field. When you create a circuit, you move the potential of the positive terminal closer to the negative terminal. If you assume that "wire" has no resistance (for simplicity) ...

3

The answer is simple, the resistor doesn't know what voltage drop to provide, and to some extent it doesn't "care" either (unless it's so high that the current fries it but that's another issue). The "voltage drop" is only governed by the potential difference created by the generator (battery or other). If the circuit is open, the electrons have no path so ...

2

Essentially, the answer to your question is yes but your equation is not quite in the general form. Typically, impedance is $$Z=R + jX$$ with $R$ being the resistance, and $X$ being the reactance which is almost the equation you show, but without the imaginary component. Specifically, $$X = \omega L - \frac{1}{\omega C}$$. What this means is that a ...

2

The battery doesn't do any significant work if there's nothing connected to it. You can think of a battery with nothing connected to it as being a battery with an extremely large resistor connected to it. That's even essentially physically accurate, instead of just being an idealization, because although the resistivity of air is extremely high, it's not ...

2

The 3 A that leaves the battery depends on two things: the amount of supply voltage, and the impedance (here resistance) that draws this current. So the second part simply says: The 3 A current that leaves the voltage source = The voltage source divided by the total resistance (which is the internal resistance, plus the equivalent of the parallel ...

2

First, the reactive power is not dissipated, but which corresponds to power delivered by the power stored in the reactive component (inductor or capacitor) during a semi-cycle; in the next half cycle, the component returns the stored energy to the source. For this to occur, the component must have the ability to store energy. In the case of a capacitor, the ...

2

Current flows in a closed loop, so $$I_c = I_r\tag1$$ There is a voltage across the capacitor due to the integrated charge, across the resistor due to the instantaneous current, and across the whole loop due to the motion of the resistor. The net voltage around the loop is zero, so $$V_c - I_r R + \mathcal{E}=0\tag2$$ The force on the wire is given by ...

2

You are massively overthinking the problem. The collector current is given (by the diagram) to be 150x the base current. The sum of base and collector current has to flow through the emitter... That's all you need to solve this. In particular, a current source will look to a circuit like "whatever resistance" it needs to be in order for the correct current ...

2

AC has each wire positive half the cycle and negative the other half. It will charge the capacitor on one half cycle and discharge it on the other half. The net charge will be zero.

2

Of course you can charge a capacitor with AC. The problem is that you keep changing how it is charged. While you apply a positive voltage to one plate, it will get a positive charge; half a cycle later, it will attempt to get a negative charge; and so it continues. The capacitor is always a little bit behind - as your AC voltage is changing, the capacitor ...

1

Like any set of equations in physics, this is true when the conditions stated apply. Here, you are using two equations: $$V(t) = RI(t)$$ when the current passes through an element that is a perfect resistor, and $$C \frac{dV}{dt} = I(t)$$ when the current passes through an element that is a perfect capacitor. So, in a simple RC circuit consisting of a ...

1

We always consider than wire has negligible resistance and if the wire is 99% copper wire it usually has resistance too small to cause drop in voltage measurable by common multimeter used in laboratory. Considering that the wire you used was such wire, between a and b, c and d and from a to + terminal and e to - terminal $P.D \approx 0V$ . Though between b ...

1

If $E$ is the emf (open-circuit voltage) of the battery and $r_s$ is the internal resistance, the equation relating the series current $I_S$ through an external resistance $R_L$ is given by: $$I_S = \frac{E}{r_s + R_L}$$ Now, the voltage across the battery terminals $V_{BAT}$ is given by $$V_{BAT} = E - I_S \cdot r_s = E\frac{R_L}{r_s + R_L}$$ So, even ...

1

Your method for evaluating the electric field assumes it is appropriate to model it as spatially constant within the wire since you're basically taking a spatial average. You'll have to decide wether or not this is accurate.

1

Well I imagine you know how to add resistors in parallel: $$R = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}}$$ Of course if you try to put a resistor with $R=0$ in there you run into $\frac{1}{0}$. You can probably take a limit and show that the equivalent resistance is $0$ if there's a zero resistor in parallel. This makes sense. If current is ...

1

One approach to gaining insight to the answer is to note that, for two parallel resistors, the equivalent resistance is $$R_{EQ} = \frac{R_1R_2}{R_1 + R_2} = \frac{R_2}{1 + \frac{R_2}{R_1}}$$ as long as either resistance is not equal to zero. Now, let $R_2$ go to almost zero while holding $R_1$ constant. When $R_2 \ll R_1$, the equivalent resistance is ...

1

The answer-as requested by HDE. So my foolish misconstruct of the electric field is that since it accelerates charged particles more or less depending on the distance from the source, it should create less motion in electrons a further distance from the source. I'm not sure why I thought resistance had much to do with it-on further analysis resistance ...

1

There is a "capacitor" of some kind holding the charge at one end of the wire - with capacitance C. It has initial charge Qo, and initial voltage Vo = Qo/C. In this relatively simple case, a discharging capacitor has an exponential rate of discharge. This website describes how to derive the exponential equation: ...

1

One equation is for resistive circuit and the other is for capacitive circuit. Two can not be merged together.

1

Yes. It depends on how you define the direction of positive voltage in your equivalent voltage source. In other words, if you switch the positions of the "+" and "-" on a typical diagram, the sign of the voltage will switch.

1

The answer depends on what you're trying to do. If you're trying to find the voltage between A and B, then note that the circuit can be redrawn as From this redrawn circuit, it's clear that the resistors need to be treated as being in series in the process of calculating the voltage between A and B. But if you're trying to find the Thévenin equivalent ...

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